Question

if tan theta =3 then evaluate 4sina-cosa/4sina+cosa

Answer 1

Answer:

divide & multiply by cos theta

Step-by-step explanation:

$\frac{4 \tan \alpha - 1}{4 \tan\alpha + 1} \\ = \frac{12 - 1}{12 + 1} \\ = \frac{11}{13}$

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Answer 2

SOLUTION

GIVEN

tan θ = 3

TO DETERMINE

$\displaystyle \sf{ \frac{4 \sin \theta - \cos \beta }{4 \sin \theta + \cos \theta } }$

EVALUATION

Here it is given that

tan θ = 3

Now

$\displaystyle \sf{ \frac{4 \sin \theta - \cos \theta }{4 \sin \theta + \cos \theta } }$

Dividing both of the numerator and denominator by cos θ we get

$\displaystyle \sf{ = \frac{4 \dfrac{\sin \theta}{ \cos \theta} - \dfrac{ \cos \theta}{ \cos \theta} }{4 \dfrac{\sin \theta}{ \cos \theta} + \dfrac{ \cos \theta}{ \cos \theta}} }$

$\displaystyle \sf{ = \frac{4 \tan \theta - 1}{4 \tan \theta + 1} }$

$\displaystyle \sf{ = \frac{12 - 1}{12 + 1} }$

$\displaystyle \sf{ = \frac{11}{13} }$

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